Question 184058
{{{(c-d)r^3-(c-d)s^3}}} Start with the given expression.



Let {{{z=c-d}}}



{{{zr^3-zs^3}}} Replace each {{{c-d}}} term with "z"




{{{z(r^3-s^3)}}} Factor out the GCF "z"




{{{z(r-s)(r^2+rs+s^2)}}} Factor {{{r^3-s^3}}} by the <a href="http://www.purplemath.com/modules/specfact2.htm">difference of cubes formula</a>. Note: let me know if you need help with this part.



{{{(c-d)(r-s)(r^2+rs+s^2)}}} Plug in {{{z=c-d}}}




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Answer:



So {{{(c-d)r^3-(c-d)s^3}}} completely factors to {{{(c-d)(r-s)(r^2+rs+s^2)}}}